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I am trying to simplify the following but I cannot. $$ \frac{\csc \alpha +\cos \alpha}{\cos \alpha - \tan \alpha - \sec \alpha} $$

Can it be simplified?

Edit

My last result is

$$ - \frac{\cos \alpha \left( 1 + \sin \alpha \cos \alpha\right)} {\sin^2 \alpha \left(1+\sin \alpha\right)} $$

I am wondering it might be a wrong question given by my student's teacher.

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    $\begingroup$ I don't think so. What did you hope it simplified to? Why do you think it's wrong? $\endgroup$ – John Doe May 17 at 7:09
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    $\begingroup$ @JohnDoe: I don't know what is the final simplification the teacher wants. Simple is too subjective word. $\endgroup$ – Gold Digging Programmer May 17 at 7:25
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If you multiply the numerator and denominator of your last expression by $4(1-\sin\alpha)$ and use the identity $\sin2\alpha=2\sin\alpha\cos\alpha$, you can rewrite your result as $$(1-\csc\alpha)(2\csc2\alpha+1).$$ This reduces the original 24 symbols to 17.

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First, let us state of the fundamentals. $$\csc \alpha = \frac{1}{\sin \alpha};\ \tan \alpha = \frac{\sin \alpha}{\cos \alpha};\ \cot \alpha = \frac{\cos\alpha}{\sin \alpha};\ \sec \alpha = \frac{1}{\cos \alpha};\ \sin^2\alpha + \cos^2\alpha = 1;$$

So, $$\frac{\csc \alpha +\cos \alpha}{\cos \alpha - \tan \alpha - \sec \alpha} \cdot 1= \frac{\frac{1}{\sin \alpha} + \cos \alpha}{\cos \alpha - \frac{\sin\alpha}{\cos\alpha} - \frac{1}{\cos\alpha}} \cdot\frac{\cos \alpha}{\cos \alpha} = \frac{\cot \alpha + \cos^2\alpha}{\cos^2\alpha - \sin\alpha - 1} = \frac{\cot\alpha + 1 -\sin^2\alpha}{1-\sin^2\alpha - \sin\alpha - 1} = \frac{\cot\alpha + 1 -\sin^2\alpha}{\left(0-\sin\alpha\right)\cdot\left(1+\sin\alpha\right)}\cdot\frac{\sin\alpha}{\sin\alpha} = \frac{\cos\alpha + \sin\alpha - \sin^3\alpha}{- \sin^2\alpha\left(1+\sin\alpha\right)} = -\frac{\cos\alpha + \sin\alpha\cos^2\alpha}{\sin^2\alpha\left(1+\sin\alpha\right)} = - \frac{\cos\alpha\left(1 + \sin\alpha\cos\alpha\right)}{\sin^2\alpha\left(1+\sin\alpha\right)}$$ Alright, a great number of further simplifications are possible from here, but what do you think will eliminate the denominator and bring the order down to one without changing the inner $\alpha$. There are 4 terms and various manipulations left.

You'll see this is actually the simplest you can get it to. It is not trivial to prove this though.

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